Problem: Let $R$ be the region enclosed by the polar curve $r(\theta)=2-2\sin(\theta)$ where $-\dfrac{2\pi}{5}\leq \theta\leq 0$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-\scriptsize\dfrac{2\pi}{5}}^{0}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$ (Choice B) B $ \int_{-\scriptsize\dfrac{2\pi}{5}}^{0}\left( 1-2\sin(\theta)+\sin^2(\theta)\right)d\theta$ (Choice C) C $ \int^{\scriptsize\dfrac{2\pi}{5}}_{0}\left( 1-2\sin(\theta)+\sin^2(\theta)\right)d\theta$ (Choice D) D $ \int^{\scriptsize\dfrac{2\pi}{5}}_{0}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=2-2\sin(\theta)}$, ${\alpha=-\dfrac{2\pi}{5}}$, and ${\beta=0}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{-\scriptsize\dfrac{2\pi}{5}}}^{{0}}\dfrac{1}{2}\left({2-2\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{-\scriptsize\dfrac{2\pi}{5}}^{0}\dfrac{1}{2}\left( 4-8\sin(\theta)+4\sin^2(\theta)\right)d\theta \\\\ &= \int_{-\scriptsize\dfrac{2\pi}{5}}^{0}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{-\scriptsize\dfrac{2\pi}{5}}^{0}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$